Đáp án đúng: B
Giải chi tiết:Gọi số mol của CH3COOH : x mol, CnHmCOOH: y mol và HOOC-COOH : z mol
CH3COOH + NaHCO3CH3COONa + CO2 + H2O (1)
x x
CnHmCOOH + NaHCO3CnHmCOONa + CO2 + H2O (2)
y y
HOOC-COOH + 2NaHCO3NaOOC-COONa + 2CO2 + 2H2O (3)
z 2z
=>\({n_{C{O_2}}} = x + y + 2z = \frac{{16,8}}{{22,4}} = 0,75\,mol\)
\({{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\frac{{{\text{21,6}}}}{{{\text{18}}}}{\text{ = 1,2}}\,{\text{mol}}\, \Rightarrow {{\text{n}}_{\text{H}}}{\text{ = }}\,{\text{2}}{{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\,{\text{2,4}}\,{\text{mol}}\)
- Bảo toàn nguyên tố oxi:
\(\begin{gathered} {{\text{n}}_{\text{O}}}{\text{ = 2}}{{\text{n}}_{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}}{\text{ + 2}}{{\text{n}}_{{{\text{C}}_{\text{n}}}{{\text{H}}_{\text{m}}}{\text{COOH}}}}{\text{ + 4}}{{\text{n}}_{{\text{HOOC - COOH}}}} \hfill \\ \,\,\,\,\,{\text{ = 2x + 2y + 4z}} \hfill \\ \,\,\,\,\,{\text{ = 2}}{\text{.0,75 = 1,5}}\,{\text{mol}} \hfill \\ \end{gathered} \)
- ĐLBTKL:
\(\begin{gathered} {m_X} = {m_C} + {m_H} + {m_O} = 44,4\,gam \hfill \\ \Rightarrow \,{n_C} = \frac{{44,4 - 2,4 - 1,5.16}}{{12}} = 1,5\,mol \hfill \\ \Rightarrow \,{n_{C{O_2}}} = {n_C} = 1,5\,mol \hfill \\ \end{gathered} \)
=> a = 1,5.44= 66 gam
Đáp án B
